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Section K The Addition Rule and the Rule of Complements In the previous section, the probabilities found were just for one event, now we will look at how to find probabilities for two or more events together, in other words compound events. (Round all answers to two or three decimal places.) A compound event is an event that is formed by combining 2 or more events. P(A or B) = P(A occurs or B occurs or both occur) – inclusive “or” P(A and B) = P(both A and B occur) Contingency Table – a table showing the distribution of one variable in rows and another in columns. Examples: 1) The following table shows the results of a survey for the income level and an individual’s favorite form of entertainment. Income Favorite Form of Entertainment Television Movies Theatre (live) Total Under $25,000 35 20 5 60 Between $25,000 and $50,000 25 18 7 50 Over $50,000 12 14 14 40 Total 72 52 26 150 A person is selected at random from this group, calculate the following probabilities: a) Find the probability that a randomly chosen individual’s favorite form of entertainment is going to the movies. P(Movies) = 𝟓𝟐 𝟏𝟓𝟎
b) P(Income is under $25,000) = 𝟔𝟎 𝟏𝟓𝟎
c) P(Income is over $50,000 or favorite form of entertainment is going to the Theatre) = 𝟓𝟐 𝟏𝟓𝟎
d) P(Income between $25,000 and $50,000 and going to the movies) = 𝟏𝟖 𝟏𝟓𝟎 = 𝟎. 𝟏𝟐 e) P(Income is over $25,000) = 𝟗𝟎 𝟏𝟓𝟎
- The following table shows the results of a survey dealing with age and gambling. A person is selected at random from this group, calculate the following probabilities: a) P(The person gambles occasionally) = 𝟔𝟎 𝟐𝟎𝟎
b) P(The person is aged between 21 and 30 or never gambles) = 𝟏𝟎𝟕 𝟐𝟎𝟎
c) P(The person is over 45 and gambles frequency) = 𝟏𝟎 𝟐𝟎𝟎
d) P(The person is over 31) = 𝟏𝟎𝟎 𝟐𝟎𝟎
e) P(The person gambles frequency or occasionally) = 𝟏𝟐𝟎 𝟐𝟎𝟎
f) P(The person is not under 20) = 𝟏𝟓𝟎 𝟐𝟎𝟎
In the above examples, since you are given a contingency table you do not need to use formulas to find probabilities, but you are not always given a contingency table so formulas are needed to find certain probabilities. The General Addition Rule For any two events A and B, P(A or B) = P(A) + P(B) – P(A and B) For any two events A and B, P(A and B) = P(A) + P(B) – P(A or B) Examples: 3 ) If P(A) = 0.35, P(B) = 0.8 and P(A and B) = 0.25. Find P(A or B). P(A or B) = 0.35 + 0.8 – 0.25 = 0. 4 ) If P(A) = 0.58, P(B) = 0.43 and P(A or B) = 0.85. Find P(A and B). P(A and B) = 0.58 + 0. 43 – 0.85 = 0. Age Gambling Frequently Occasionally Never Total Under 20 12 18 20 50 21 – 30 10 17 23 50 31 – 45 28 15 7 50 Over 45 10 10 30 50 Total 60 60 80 200
Complement If A is any event, the complement of A is the event that A does not occur. The complement of A is denoted AC. Note: P(A) + P(AC) = 1, so P(AC) = 1 – P(A) Example:
- If P(A) = 0.25 and P(B) = 0.45. Find P(AC) and P(BC). P(AC) = 1 – 0.25 = 0.75 and P(BC) = 1 – 0.45 = 0. More examples:
- A survey of type of accommodation a person lives in resulted in the following table: A person is selected at random. Find the following probabilities: a) P(the person lives in a Condo) = 𝟐𝟕𝟗 𝟏𝟕𝟑𝟔
b) P(the person lives in a House) = 𝟒𝟔𝟖 𝟏𝟕𝟑𝟔
c) P(the person lives in an apartment or a townhouse) = 𝟔𝟒𝟔+𝟑𝟒𝟑 𝟏𝟕𝟑𝟔
𝟗𝟖𝟗 𝟏𝟕𝟑𝟔
Type of Accommodation Frequency House 468 Condo 279 Apartment 646 Townhouse 343 Total 1736
AC
A
- Let B be the event that a car brought in for service needs new brakes and let S be the event the car needs new struts. Suppose that P(B) = 0.20, P(S) = 0.15 and P(B and S) = 0.05. a) Find the probability the car needs brakes or struts or both. P(B or S) = 0.20 + 0.15 – 0.05 = 0. b) Find the probability the car does not need new brakes. P(BC) = 1 – 0.20 = 0.
- Last semester at Mercer, 250 students enrolled in both MAT125 and ENG101. Of these students 38 earned an A in statistics, 50 earned an A in English and 20 earned an A in both statistics and English. a) Find the probability a randomly chosen student earned an A in MAT125 or ENG101 or both. P(M) = 𝟑𝟖 𝟐𝟓𝟎
= 𝟎. 𝟏𝟓𝟐 P(E) =
𝟓𝟎 𝟐𝟓𝟎 = 𝟎. 𝟐𝟎 P(M and E) = 𝟐𝟎 𝟐𝟓𝟎
so P(M or E) = 0.152 + 0.20 – 0.08 = 0. b) Find the probability a randomly chosen student did not earn an A in MAT125. P(MC) = 1 – 0.152 = 0.
- In a BIO103: Anatomy and Physiology class there were 40 students. 23 were females and 17 were males. Three males and six females earned an A in the course. A student is chosen at random from the class. a) Find the probability the student is a male. P(male) = 𝟏𝟕 𝟒𝟎
b) Find the probability the student earned an A in the course. P(earned an A) = 𝟑+𝟔 𝟒𝟎
𝟗 𝟒𝟎
c) Find the probability the student is male and earned an A. P(male and A) = 𝟑 𝟒𝟎
d) Find the probability the student is male or earned an A. P(male or A) = P(male) + P(A) – P(male and A) = 0.425 + 0.225 – 0.075 = 0. e) Find the probability the student did not earn an A. P(AC) = 1 – 0.225 = 0.
